Question: Simplify and expand the following expression: $ \dfrac{2}{4n + 4}- \dfrac{2}{n + 8}+ \dfrac{n}{n^2 + 9n + 8} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{2}{4n + 4} = \dfrac{2}{4(n + 1)}$ We can factor the quadratic in the third term: $ \dfrac{n}{n^2 + 9n + 8} = \dfrac{n}{(n + 1)(n + 8)}$ Now we have: $ \dfrac{2}{4(n + 1)}- \dfrac{2}{n + 8}+ \dfrac{n}{(n + 1)(n + 8)} $ The least common multiple of the denominators is: $ 4(n + 1)(n + 8)$ In order to get the first term over $4(n + 1)(n + 8)$ , multiply by $\dfrac{n + 8}{n + 8}$ $ \dfrac{2}{4(n + 1)} \times \dfrac{n + 8}{n + 8} = \dfrac{2(n + 8)}{4(n + 1)(n + 8)} $ In order to get the second term over $4(n + 1)(n + 8)$ , multiply by $\dfrac{4(n + 1)}{4(n + 1)}$ $ \dfrac{2}{n + 8} \times \dfrac{4(n + 1)}{4(n + 1)} = \dfrac{8(n + 1)}{4(n + 1)(n + 8)} $ In order to get the third term over $4(n + 1)(n + 8)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{n}{(n + 1)(n + 8)} \times \dfrac{4}{4} = \dfrac{4n}{4(n + 1)(n + 8)} $ Now we have: $ \dfrac{2(n + 8)}{4(n + 1)(n + 8)} - \dfrac{8(n + 1)}{4(n + 1)(n + 8)} + \dfrac{4n}{4(n + 1)(n + 8)} $ $ = \dfrac{ 2(n + 8) - 8(n + 1) + 4n} {4(n + 1)(n + 8)} $ Expand: $ = \dfrac{2n + 16 - 8n - 8 + 4n}{4n^2 + 36n + 32} $ $ = \dfrac{-2n + 8}{4n^2 + 36n + 32}$ Simplify: $ = \dfrac{-n + 4}{2n^2 + 18n + 16}$